Why is the differential operator equal to an integer in the case of trignometric equations?

by GenRincewind   Last Updated January 19, 2018 18:20 PM

In this website, http://www.codecogs.com/library/maths/calculus/differential/the-d-operator.php, the following problem and solution is given:


Find the Particular Integral of $$\frac{d^2y}{dx^2} - 5\,\frac{dy}{dx} + 6y = \sin\,2x$$


This can be re-written as $$ y = \frac{1}{D^2 - 5D + 6}\sin\,2x $$

Using equation 1 we can put $D^2 = -4$

Therefore: $$y = \frac{1}{-\,4\;-\,5D + 6}\sin\,2x =\frac{1}{2 - 5D}\sin\,2x$$

If we multiply the top and bottom of this equation by 2 + 5D

$$y =\frac{2 + 5D}{4 - 25D^2}\sin\,2x$$

But $D^2=-4$

Therefore $$ y =\frac{2 + 5D}{104}\sin\,2x = \frac{1}{104}\left(2\sin\,2x + 5D\sin\,2x \right)$$ But since $D\sin\,2x = 2\cos\,2x$ $$y = \frac{1}{104}\left(2\sin\,2x + 10\cos\,2x \right) $$

The biggest problem I have with this is the line, $D^2 = -4$.

So far as I understand, $D$ is an operator on functions, and $D^2$ represents the double application of that operator. So how, can it be equal to -4? How can an operator on functions be equal to an integer? They're two things of fundamentally different types, how can they be equal?

Answers 1

You can interpret $D^2=-4$ and $D^2=-4I$, where $I$ is the identity operator; that's no big deal. If anything, you should be more worried about "dividing" by an operator.

To expand a bit about what's going on in that solution, the author knows that the solution will be a linear combination of $\sin(2x)$ and $\cos(2x)$. For those two functions (and their linear combinations), $D^2=-4I$.

Note that guessing all those manipulations is harder than postulating that the particular solution is of the form $\alpha\sin(2x)+\beta\cos(2x)$ and substitute in the equation to find $\alpha$ and $\beta$ (what is usually known as Undetermined Coefficients.

Martin Argerami
Martin Argerami
January 19, 2018 17:56 PM

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